Z8 is not an algebra group: Difference between revisions

Latest revision as of 22:03, 16 August 2012

Statement

The group cyclic group:Z8, defined as the cyclic group of order $2^{3} = 8$ , is not an algebra group.

Related facts

Facts used

Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size

Proof

By Fact (1), if $Z / 8 Z$ is an algebra group over $F_{2}$ , it must be isomorphic to a subgroup of $U T (4, p)$ . However, $U T (4, p)$ has exponent 4, so $Z / 8 Z$ , which has exponent 8, cannot be isomorphic to a subgroup of it.

References

MathOverflow question: p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra

Jack Schmidt's answer sketches the above proof.

@@ Line 12: / Line 12: @@
 * [[Cyclic group of prime-square order is not an algebra group for odd prime]]
+==Facts used==
+# [[uses::Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size]]
 ==Proof==
-Note that, for a 2-group, being an algebra group is equivalent to being an algebra group over <math>\mathbb{F}_2</math>. So, we will prove that the group is an algebra group over <math>\mathbb{F}_p</math>.
+By Fact (1), if <math>\mathbb{Z}/8\mathbb{Z}</math> is an algebra group over <math>\mathbb{F}_2</math>, it must be isomorphic to a subgroup of <math>UT(4,p)</math>. However, <math>UT(4,p)</math> has exponent 4, so <math>\mathbb{Z}/8\mathbb{Z}</math>, which has exponent 8, cannot be isomorphic to a subgroup of it.
-Suppose <math>F = \mathbb{F}_2</math> and <math>N</math> is a nilpotent associative algebra over <math>F</math> such that the algebra group of <math>N</math>, which we denote <math>G</math>, is isomorphic to <math>\mathbb{Z}/8\mathbb{Z}</math>. <math>N</math> is three--dimensional over <math>F</math>, so the unitization of <math>N</math>, i.e., the algebra <math>N + F</math>, is four-dimensional. The action of <math>G</math> by left multiplication makes <math>G</math> a subgroup of <math>GL(4,F) = GL(4,2)</math> and hence of the [[Sylow subgroup]] [[UT(4,2)]] of upper triangular unipotent matrices. However, <math>UT(4,2)</math> has exponent 4, so <math>G</math>, which has exponent 8, cannot be isomorphic to a subgroup of it.
 ==References==
 * {{mathoverflow|number=68101|title=p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra|details = Jack Schmidt's answer sketches the above proof.}}